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 Author Topic: Conservation of Angular momentum and 3D circular motion  (Read 19788 times) 0 Members and 2 Guests are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: March 12, 2010, 11:06:08 pm » posted from:Taipei,T\'ai-pei,Taiwan

A particle with mass $m$ is moving with constant speed $v$ along a circular orbit (radius $r$).
The centripetal force $F=m\frac{v^2}{r}$ is provided by gravitation force from another mass $M=F/g$.
A string is connected from mass m to the origin then connected to mass $M$.
Because the force is always in the $\hat{r}$ direction, so the angular momentum $\vec{L}=m\,\vec{r}\times \vec{v}$ is conserved. i.e. $L=mr^2\omega$ is a constant.

For particle with mass m:

$m \frac{d^2r}{dt^2}=m\frac{dv}{dt}= m \frac{v^2}{r}-Mg=\frac{L^2}{mr^3}- Mg$
$\omega=\frac{L}{mr^2}$

The following is a simulation of the above model.

You can change the mass M or the radius r with sliders.
The mass M also changed to keep the mass m in circular motion when you change r.
However, if you change mass M , the equilibrium condition will be broken.

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macfamous
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 « Embed this message Reply #1 on: March 19, 2010, 10:53:15 am » posted from:Yogyakarta,Yogyakarta,Indonesia

there is more simple formula for this simulation
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Fu-Kwun Hwang
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 « Embed this message Reply #2 on: March 19, 2010, 01:22:45 pm » posted from:Taipei,T\'ai-pei,Taiwan

Quote
there is more simple formula for this simulation

Could you explain what do you mean in detail?
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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #3 on: July 07, 2010, 12:32:23 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

question:
how is the tension in the string calculated?

My understanding:
T = m*v*v/r; // added lookang for tension only work for circular motion

my hypothesis:
T1 = m*v*v/r + m*dvr/dt ? // radial acceleration due to tangential change in velocity and radial acceleration due to radial change in velocity ?

where dvr/dt = (cst*cst/(m*r*r*r)-M*g)/m

My equation:
T1 = m*v*v/r +(cst*cst/(m*r*r*r)-M*g) ;
can help take a look if my equation of Tension is correct ?
did i get the generalized equation for tension in any elliptical motion?
the logic seems correct and i implement the equation looks correct. is it correct conceptually in the absence of frictional forces
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Fu-Kwun Hwang
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 « Embed this message Reply #4 on: July 07, 2010, 06:05:49 pm » posted from:Taipei,T\'ai-pei,Taiwan

If the mass is in circular motion, then the tension $T=m\frac{v^2}{r}$.

What if we increase the tension so that $T>m\frac{v^2}{r}$
Will the mass moving inward($\frac{dv_r}{dt}<0$) or moving outward($\frac{dv_r}{dt}>0$)?

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lookang
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 « Embed this message Reply #5 on: July 07, 2010, 06:16:08 pm »

If the mass is in circular motion, then the tension $T=m\frac{v^2}{r}$.

What if we increase the tension so that $T>m\frac{v^2}{r}$.
Will the mass moving inward($\frac{dv_r}{dt}<0$) or moving outward($\frac{dv_r}{dt}>0$)?

I see I got the sign convention wrong!
I forgot the convention for r unit vector is positive outward.
Will change it to
T1 = m*v*v/r -(cst*cst/(m*r*r*r)-M*g) ;
thanks !!!

your masterful questioning allows to think and figure out the answer myself.